Question

8 sin^2x-2cosx-5=0 Пожалуйста помогите, заранее спасибо!

Answer 1

8sin^2x-2cosx-5=0,

8(1-cos^2x)-2cosx-5=0,

-8cos^2x-2cosx+3=0,

8cos^2x+2cosx-3=0,

cosx=t,

8t^2+2t-3=0,

D=100,

t1=-3/4,

t2=1/2,

 

cosx=-3/4,

x=-+arccos(-3/4) +2pi*k, kєZ,

x=-+(pi-arccos(3/4)) +2pi*k, kєZ,

 

cosx=1/2,

x=-+arccos(1/2)+2pi*k, kєZ,

x=-+pi/3 + 2pi*k, kєZ

 

Answer 2

8 sin^2x-2cosx-5=0

8cosx^2+2cosx-3=0

 8x2 + 2x - 3 = 0
D = b2 - 4ac
D = 4 + 96 = 100 = 10^2

x1,2 = -b ± √D/2a
x1 = -2 + 10/16 = 8/16 = 1/2
x2 = -2 - 10/16 = - 12/16 = - 3/4
Ответ: x1 = 1/2 ; x2 = - 3/4